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8.01 Solving right triangles

Lesson

Finding a missing side

As we recall from Geometry, there are $3$3 trigonometric ratios that relate an angle and sides together.  They are:

Trigonometric ratios

       $\sin\theta$sinθ = $\frac{Opposite}{Hypotenuse}$OppositeHypotenuse  

       $\cos\theta$cosθ = $\frac{Adjacent}{Hypotenuse}$AdjacentHypotenuse 

       $\tan\theta$tanθ = $\frac{Opposite}{Adjacent}$OppositeAdjacent 

If we know any $2$2 parts of a right triangle, whether that's $2$2 side lengths, or an angle and a side length we can then find any other part of that triangle using these trigonometric ratios.  

 

Finding side lengths

If we know $2$2 sides and want to find the third, we would use the Pythagorean formula

Pythagorean theorem

$a^2+b^2=c^2$a2+b2=c2

If we know $1$1 side length and an angle, we would use one of the trigonometric ratios.

The most common mistake is when the wrong ratio is used.  We have to remember the ratios and the sides that apply to those ratios.  For most students the mnemonic SOHCAHTOA can be a great help.

Worked examples

question 1

In the given triangle $\theta=25^\circ$θ=25° and the hypotenuse measures $12.6$12.6. Solve for the length $b$b. Round to the tenth place.

Think: We need to identify the sides we have and want with respect to the angle given.  Here I can see that we have the hypotenuse (H) and we want $b$b, which is opposite (O) the angle.  This means I have OH - so the trig ratio I need to use here is sine.  

Do:

$\sin\theta$sinθ $=$= $\frac{O}{H}$OH
$\sin25^\circ$sin25° $=$= $\frac{b}{12.6}$b12.6
$b$b $=$= $12.6\times\sin25^\circ$12.6×sin25°
$b$b $=$= $5.32$5.32

 


question 2

Find the length of the hypotenuse ($c$c) in the diagram, where the angle $36^\circ$36° and the side length of $4.8$4.8 are given. Round to $2$2 decimal places.

Think: We need to identify the sides we have and want with respect to the angle given.  Here I can see that we want the hypotenuse (H) and we have a side length of $4.8$4.8, which is adjacent (A) the angle.  This means I have AH - so the trig ratio I need to use here is cosine. 

Do:

$\cos\theta$cosθ $=$= $\frac{A}{H}$AH
$\cos36^\circ$cos36° $=$= $\frac{4.8}{c}$4.8c
$c$c $=$= $\frac{4.8}{\cos36^\circ}$4.8cos36°
$c$c $=$= $5.93$5.93

question 3

Find the length of the unknown side, when the angle is $66^\circ$66° and the indicated side length is $7.3$7.3. Answer rounded to the tenths place.

Think: We need to identify the sides we have and want with respect to the angle given.  Here I can see that we want the adjacent side (A) and we have a side length of $7.3$7.3, which is opposite (O) the angle.  This means I have OA - so the trig ratio I need to use here is tangent. 

Do:

$\tan\theta$tanθ $=$= $\frac{O}{A}$OA
$\tan66^\circ$tan66° $=$= $\frac{7.3}{a}$7.3a
$a$a $=$= $\frac{7.3}{\tan66^\circ}$7.3tan66°
$a$a $=$= $3.3$3.3


Practice questions

QUESTION 4

Find the value of $f$f, correct to two decimal places.

A right triangle with an interior angle of $25$25 degrees. The side adjacent to the $25$25-degree angle has a length of $11$11 mm and its opposite side measures f mm.

Question 5

Find the value of $h$h, correct to two decimal places.

A right triangle whose measure of hypotenuse is unknown and is labeled $h$h m. There is an indicated interior angle that measures $42$42 degrees and the side opposite to this angle measures $11$11 $m$m long. The side adjacent to $42$42 degrees is not labeled.

Question 6

Find the value of $x$x, the side length of the parallelogram, to the nearest centimeter.

A parallelogram with its angle at upper-right corner labeled as 52 degrees, indicating its measure. Other angles of the parallelogram are not labeled. Its top side is labeled 32 cm, indicating its length. Its left side is labeled $x$x cm, indicating its unknown length. A perpendicular internal segment is drawn from the upper-left corner to the bottom side of the parallelogram, thus creating a right triangle on the left side. The bottom side of the parallelogram, which is parallel to the top side as indicated by the double arrowheads, is cut into two parts: the left part with no measurement indicated and the right part which measures 11 cm, as labeled.

In the triangle, the sides are the internal segment, the left side of the parallelogram, and the left part of the bottom side of the parallelogram. A square symbol is shown at the corner where the internal segment and the bottom side intersect to indicate that it is a right angle. The left side of the parallelogram labeled $x$x cm acts as the hypotenuse of the triangle. The internal segment is the vertical leg of the triangle. The left part of the bottom side of the parallelogram acts as horizontal leg of the triangle.

 

Finding a missing angle

We know how to find the length of unknown sides of right triangles. Now, we can use the same trigonometric ratios to find the unknown angles.  To do this we need any 2 of the side lengths.

Remember

Also,

  • label the sides as O, A or H with respect to the position of the angle you want to find
  • identify the appropriate trigonometric ratio that applies [either sine (sin), cosine (cos) or tangent (tan)]
  • using algebra, solve the equation for the unknown, (write the rule, fill in what you know, then solve using inverse operations)
  • reflect and check (do a quick check on your calculator to confirm your answer is correct)

Worked examples

question 7

What is value of $\theta$θ? Round to the hundredth place.

Think: Label the sides as O, A or H with respect to the position of the angle. Identify the appropriate ratio that uses O and H.  

For this question it will be $\sin$sin. Isolate the unknown using algebraic techniques. 

Do:

$\sin\theta$sinθ $=$= $\frac{O}{H}$OH
$\sin\theta$sinθ $=$= $\frac{5}{8}$58
$\theta$θ $=$= $\sin^{-1}\left(\frac{5}{8}\right)$sin1(58)
$\theta$θ $=$= $38.68^\circ$38.68°
question 8

Find the value of the angle indicated. Round to $2$2 decimal places.

 

Think: Label the sides as O, A or H with respect to the position of the angle. Identify the appropriate ratio that uses O and H. For this question it will be $\tan$tan. Isolate the unknown using algebraic techniques.  

Do: 

$\tan\theta$tanθ $=$= $\frac{O}{A}$OA
$\tan\theta$tanθ $=$= $\frac{14.77}{12.24}$14.7712.24
$\theta$θ $=$= $\tan^{-1}\left(\frac{14.77}{12.24}\right)$tan1(14.7712.24)
$\theta$θ $=$= $50.35^\circ$50.35°

 

Practice questions

Question 9

Find the value of $x$x to the nearest degree.

A right triangle with vertices labeled A, B and C. Vertex A is at the top, B at the bottom right, and C at the bottom left. A small square at vertex A indicates that it is a right angle. Side interval(BC), which is the side opposite vertex A, is the hypotenuse and is marked with a length of 25. The angle located at vertex B is labeled x. Side interval(AB), descending from the right angle at vertex A to vertex B, is  marked with a length of 7, and is adjacent to the angle x. Side interval(AC) is opposite the angle x.

Question 10

Consider the given figure.

  1. Find the unknown angle $x$x, correct to two decimal places.

  2. Find $y$y, correct to two decimal places.

  3. Find $z$z correct to two decimal places.

Question 11

The person in the picture sights a pigeon above him. If the angle the person is looking at is $\theta$θ, find $\theta$θ in degrees.

The image depicts a right-angled triangle formed by dashed lines. A person in blue standing on the ground has a horizontal line of sight that represents the base of the triangle which measures 12m indicating the base length. From the base of the triangle, the vertical height of the pigeon above the man is 27m. The angle between the base and the hypotenuse is marked with a theta (θ) symbol, indicating an unknown angle to be potentially calculated.

  1. Round your answer to two decimal places.

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